Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) Members Public

My solutions for Burton's Elementary Number Theory Problems 4.2 (7th Edition)

Ran
Ran
Math

Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) - Q18 Members Public

My Solution for "If $a \equiv b \pmod {n_{1}}$ and $a \equiv c \pmod {n2}$, prove that $b \equiv c \pmod {n}$, where the integer $n = gcd(n_{1}, n_{2})$."

Ran
Ran
Math

Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) - Q17 Members Public

My Solution for "Prove that whenever $ab \equiv cd \pmod {n}$ and $b \equiv d \pmod {n}$, with $gcd(b, n) = 1$, then $a \equiv c \pmod {n}$."

Ran
Ran
Math

Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) - Q16 Members Public

My Solution for "Use the theory of congruences to verify that $$89 \mid 2^{44} - 1 \qquad \text{and} \qquad 97 \mid 2 ^{48} - 1$$"

Ran
Ran
Math

Techniques I learned from Playing Sekiro Members Public

Struggling with Sekiro? Try the techniques I realized while playing Sekiro. From being stuck on Genichiro for four nights to mastering these techniques and defeating Isshin in just 3 hours.

Ran
Ran
Sekiro

Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) - Q15 Members Public

My Solution for "Establish that if $a$ is an odd integer, then for any $n \geq 1$$$a^{2^{n}} \equiv 1 \pmod {2^{n + 2}}$$ [Hint: Proceed by induction on $n$.]"

Ran
Ran
Math

Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) - Q14 Members Public

My Solution for "Give an example to show that $a^{k} \equiv b^{k} \pmod {n}$ and $k \equiv j \pmod {n}$ need not imply that $a^{j} \equiv b^{j} \pmod {n}$."

Ran
Ran
Math

Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) - Q13 Members Public

My Solution for "Verify that if $a \equiv b \pmod {n_{1}}$ and $a \equiv b \pmod {n_{2}}$, then $a \equiv b \pmod {n}$, where the integer $n = lcm(n_{1}, n_{2})$. Hence, whenever $n_{1}$ and $n_{2}$ are relatively prime, $a \equiv b \pmod{n_{1}n_{2}}$."

Ran
Ran
Math